Solutions · Lesson 4 of 11
Vapour Pressure of Volatile Liquid Mixtures
“Use liquid composition to calculate equilibrium partial pressures, total pressure and vapour composition.”
• Explain equilibrium vapour pressure and distinguish it from evaporation rate. • Apply Raoult’s law to an ideal mixture of volatile liquids. • Combine Raoult’s and Dalton’s laws to obtain vapour composition. • Interpret pressure–composition graphs and compare Henry and Raoult limits.
A liquid and its vapour reach a dynamic balance
In a closed vessel, molecules escape a liquid while vapour molecules return to it. Initially, evaporation may dominate. As vapour accumulates, condensation becomes more frequent until the two rates are equal. At a fixed temperature the resulting equilibrium pressure is the vapour pressure. Its value reflects how readily the molecules transfer into the vapour, rather than merely how much liquid surface is exposed. Surface area can affect equilibration time, but not the final equilibrium vapour pressure of a pure liquid when both phases remain.
If both components of a liquid mixture are volatile, both normally enter the vapour. Label the liquid mole fractions x₁ and x₂, with x₁+x₂=1. The vapour has its own mole fractions y₁ and y₂; they need not equal the liquid fractions. A more volatile pure liquid has a higher pure-component vapour pressure at the same temperature.
Raoult’s law connects liquid fraction to partial pressure
In an ideal mixture, replacing some molecules of component 1 by similar interacting molecules of component 2 reduces component 1’s equilibrium vapour contribution in proportion to its liquid fraction. The pure liquid is the endpoint x₁=1, giving p₁=p₁°. At x₁=0, no component 1 remains, and p₁=0. Thus its pressure–composition line must connect those endpoints.
The second equality substitutes the Raoult expressions. The third uses x₁=1−x₂, and expanding brackets gives the final straight-line form. At x₂=0, total pressure is p₁°; at x₂=1 it is p₂°. If p₂° exceeds p₁°, total pressure rises as component 2 is added. The slope is p₂°−p₁° and the intercept is p₁°. The result lies between the two pure-component pressures for every composition in an ideal binary mixture.
Problem
At a fixed temperature, pure liquids A and B have vapour pressures 450 and 700 mm Hg. An ideal mixture has total pressure 600 mm Hg. Find its liquid composition.
- 1.Use x(A)=1−x(B). The pressure equation becomes 600=450(1−x(B))+700x(B).
- 2.Collect terms: 600−450=(700−450)x(B). Hence x(B)=150/250=0.600 and x(A)=0.400.
- 3.Both fractions lie between zero and one and sum to one. Total pressure is between the pure pressures, as required.
Vapour composition is obtained from partial pressures
For the A–B mixture above, p(A)=0.400×450=180 mm Hg and p(B)=0.600×700=420 mm Hg. Dividing by 600 gives y(A)=0.300 and y(B)=0.700. Vapour is enriched in B relative to the liquid, because B contributes more vapour pressure per mole fraction. For an ideal binary mixture with both components present and p(B)°>p(A)°, y(B)>x(B). This is one reason distillation can change composition.
Problem
Assume ideal behaviour at 298 K. Mix 25.5 g CHCl₃ and 40.0 g CH₂Cl₂. Pure vapour pressures are 200 and 415 mm Hg; molar masses are 119.5 and 85 g mol⁻¹. Find total pressure and vapour composition.
- 1.Convert masses separately: n(CHCl₃)=25.5/119.5=0.2133891 mol; n(CH₂Cl₂)=40/85=0.4705882 mol.
- 2.Liquid fractions are x(CH₂Cl₂)=0.6880173 and x(CHCl₃)=0.3119827. Mass fractions would be inappropriate here.
- 3.Partial pressures are 415×0.6880173=285.5272 mm Hg and 200×0.3119827=62.3965 mm Hg. Total =347.9237 mm Hg.
- 4.Vapour fractions are y(CH₂Cl₂)=285.5272/347.9237=0.8206603 and y(CHCl₃)=0.1793397. The more volatile dichloromethane is enriched.
- 5.These are model predictions for the specified ideal assumption; a real liquid pair may require measured deviations.
Problem
A mixture has 100 g A of molar mass 140 and 1000 g B of molar mass 180 g mol⁻¹. Pure B pressure is 500 torr and ideal total pressure is 475 torr. Find pure A pressure and A partial pressure.
- 1.Amounts are n(A)=100/140=5/7 mol and n(B)=1000/180=50/9 mol. Thus x(A)=9/79 and x(B)=70/79.
- 2.p(B)=(70/79)×500=443.0380 torr. Therefore p(A)=475−443.0380=31.9620 torr.
- 3.Raoult’s law gives p(A)°=p(A)/x(A)=31.9620/(9/79)=280.5556 torr.
- 4.The pure A pressure is not its partial pressure. Its smaller pure pressure explains why the total is below 500 torr.
Henry’s law and Raoult’s law: similar algebra, different reference states
Both expressions have the form p proportional to liquid mole fraction. For a dilute solute, the proportionality constant is its Henry constant. For an ideal solution component referenced to its pure liquid, the constant is its pure-liquid vapour pressure. If K_H equals that pure pressure, the two algebraic forms coincide. In real mixtures Henry behaviour describes the dilute-solute limit, while Raoult behaviour is approached by the solvent near its pure-component limit. The shared linear form does not prove that every real mixture obeys Raoult’s law at all compositions.
Do not insert gas-phase mole fraction y into p=xp°. Raoult’s law uses liquid x. Dalton’s law p=yp_total uses vapour y. First calculate component amounts in the liquid, then partial pressures, then vapour fractions.
Quick Quiz
Quiz
Which condition defines equilibrium vapour pressure?
Explanation: The rates balance while molecular exchange continues; the two phases need not have the same composition.
Pure pressures are 40 and 80 kPa. What is total pressure at equal liquid mole fractions in an ideal mixture?
Explanation: p=0.5×40+0.5×80=60 kPa.
Which expression calculates vapour composition?
Explanation: Dalton’s law gives pᵢ=yᵢp_total, hence yᵢ=pᵢ/p_total.
For an ideal binary liquid with p(B)°>p(A)° and both components present, which is true?
Explanation: The more volatile component B has a larger share of the vapour than of the liquid. Total ideal pressure remains between the endpoint pressures.
Which pressure is multiplied by xᵢ in Raoult’s law?
Explanation: The proportionality constant is the same component’s pure-liquid vapour pressure at the specified temperature.
Practice Problems
- An ideal mixture contains 26.0 g heptane and 35.0 g octane at 373 K. Use molar masses 100 and 114 g mol⁻¹ and pure pressures 105.2 and 46.8 kPa. Find total pressure.
- Mix 80 g benzene and 100 g toluene at 300 K. Use molar masses 78 and 92 g mol⁻¹ and pure pressures 50.71 and 32.06 mm Hg. Find the benzene vapour fraction assuming ideality.
- At x(B)=0.25, pure pressures are p(A)°=100 and p(B)°=200 kPa. Find both partial pressures and vapour fractions.
- Describe the pressure–composition graph for the preceding mixture as x(B) goes from zero to one.
- Why can the expression p=K_Hx not establish ideality over a whole composition range?
Amounts are 0.260 mol heptane and 35/114=0.3070175 mol octane. x(heptane)=0.260/0.5670175=0.4585396. Total pressure =0.4585396×105.2+0.5414604×46.8=73.5787 kPa≈73.6 kPa. It lies between 46.8 and 105.2 kPa. Convert masses to moles first because Raoult’s law uses liquid mole fractions, then add both component contributions.
Key Takeaways
• Vapour pressure is an equilibrium property at a specified temperature. • Raoult’s law for an ideal mixture uses liquid mole fractions. • The total pressure is the sum of partial pressures. • Ideal total pressure varies linearly between pure-component endpoint pressures. • Dalton’s law converts partial pressures into vapour mole fractions. • Vapour is enriched in the more volatile component of an ideal binary mixture. • Henry and Raoult relations share a linear form but use different reference limits.